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### Applications of Rational Expressions

- When working on a word problem the unknown could end up in the denominator. These often lead to rational expression.
- Some common word problems that lead to rational expressions are ones that involve motion or work.
- If a job can be completed in t units of time, then the rate of work is given by 1/t.
- By multiplying the rate by the amount of time, we can figure out how much of a job has been completed at any given time.
- From working with motion we know that distance = rate x time. This leads to other equations such as rate = distance/time and time = distance/rate

### Applications of Rational Expressions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Objectives 0:07
- Applications of Rational Expressions 0:27
- Work Problems
- Example 1 2:58
- Example 2 6:45
- Example 3 13:17
- Example 4 16:37

### Algebra 1 Online Course

### Transcription: Applications of Rational Expressions

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at applications of rational equations.*0002

*In some of the examples I have cooked up we will look at how some examples involves a just numbers.*0009

*There are ones that involve motion and some of my favorite ones that involve work.*0015

*This one is going to be a little bit trickier to think about, but once you get the process done*0020

*you will see that the work problems are not so bad.*0024

*Depending on the unknowns in the problem and depending on how we should go ahead and package everything up*0031

*there could be a few situations that actually lead to rational expressions.*0037

*Think about those fractions or where an unknown ends up on the denominator.*0041

*What we want to do is be able to solve these using many of the techniques that we have picked up for rational equations*0050

*and see how we can recognize these in various different situations, such as numbers, motion, and especially work.*0056

*The work problems are kind of unique and that you want to know how to represent the situation.*0067

*If you know how long it takes to complete an entire job then you know the rate of work is given by the following formula 1/T.*0075

*The way you can read this is by using that T for the amount it takes to do that one job.*0084

*Here is a quick example.*0090

*Let us say it takes Betty 7 hours to paint her entire room.*0092

*Well, that means that every single hour 1/7 of the room is going to be painted.*0098

*We are going to make a little chart for just keeping track of everything.*0103

*Maybe x will be my time and let us say how much of the room has been painted so far.*0109

*One hour, two hours, three hours and make sure you jump all the way to 7.*0120

*After she is in the room for 7 hours, she will have the entire room painted, the whole thing.*0127

*If you scale it back, what if she is only working for one hour then only 1/7 of the room is painted.*0135

*If she is in there for two hours, 2/7 of the room is painted and if she is in there for three 3/7.*0144

*You can see that we are just incrementing this thing by exactly 1/7 every time.*0151

*We can make some adjustments to our formula here and say that T would be the amount it takes to do that one for the person*0156

*and may be multiply it by x, that would represent how long they have been doing that particular job.*0165

*Watch for that to play a key component with our work problem in just a bit.*0173

*Let us first see our example of numbers and just see something where our variable ends up in the denominator.*0180

*In this one we have a certain number and we are going to add it to the numerator and subtract it from the denominator of 7/3.*0186

*The result equals the reciprocal of 7/3 and we are interested in finding that number.*0194

*Let us first write down our unknown.*0202

*x is the unknown number.*0205

*We construct a model situation here.*0217

*Take that unknown number and add it to the numerator, but subtract it from the denominator of 7/3.*0219

*Here is 7/3, so we are adding it to the numerator and subtract it from the denominator.*0227

*The result equals the reciprocal of 7/3, that is like 7/3 but we flipped it over.*0238

*This will be our rational inequality here.*0245

*What we have to do is work on solving it.*0249

*To solve many of our rational equalities we work on finding a least common denominator*0253

*which I can see for this one will be 3 - x and 7.*0261

*Let us give that missing piece to each of the fractions.*0268

*Here is my original (x + x) (3 – x) = 3/7 let us use some extra space in here.*0272

*The one on the left, it could use 7, let us put that in there.*0286

*The one on the right it is missing the 3 – x.*0294

*At that point, the denominators will be exactly the same.*0302

*We will just go ahead and focus on the tops of each of these.*0307

*7 × 7 + x = 3, 3 – x.*0312

*Continuing on and solving for our x we will go ahead and distribute our 7 and 3.*0321

*That will give us 49 7x = 9 - 3x.*0328

*Moving along pretty good.*0336

*Let us go ahead and add 3x to both sides giving us a 49 + 10x = 9.*0338

*We will go ahead and subtract 49 from both sides.*0351

*I have x = 10x is equal to -40.*0359

*Dividing both sides by 10, I have that x = -4.*0368

*Just like when we are working with equations, it has to make sense in our original.*0374

*Looking at the original rational expression I have a restricted value of 3 and I know that it is not 3 that will make the bottom 0.*0379

*I do not get any restricted values from the other faction because it is simply always 7 on the bottom.*0388

*Since the -4 is not 3, I'm going to keep it as a valid solution.*0394

*That one looks good.*0402

*Let us look at one that involves motion.*0407

*We will set these up using a table and also use that same idea to help us organize this information.*0409

*A boat can go 10 miles against a current in the same time it can go 30 miles with the current.*0416

*The current flows at 4 mph, find the speed of the boat with no current.*0422

*We have an interesting situation.*0428

*We have a boat that looks something like this and we have the flow of the river.*0430

*Now in one situation, it is fighting against the current, and the way you want to think of that in relation to its speed*0439

*is that the speed of the river is taking away some of the speed of the boat.*0447

*You will see a subtraction process.*0452

*If the boat is going in the same direction of the river, they are both helping each other out*0456

*and you will see an addition problem with both of their speed.*0461

*You will know they are both helping each other out.*0464

*Let us see if we organize this information so I can see how to connect it.*0468

*We need to think of two different situations.*0492

*We are either going against, or we are going with the river.*0495

*We will look at the rate, the time, and the distance.*0502

*This will help us keep track of everything.*0511

*And of course we are leaving unknown in here.*0513

*Since we are finding the speed of the boat with no current, let us set that as our unknown.*0516

*x is the speed of the boat with no current.*0522

*I think we have a good set up and we can start organizing our information.*0540

*In the first bit of this problem we know that it can go 10 miles where it is going against the current.*0545

*That is its distance.*0553

*It went 10 miles when it is going against the current.*0554

*It can do that in the same time it can go 30 miles with the current.*0558

*A little bit of different information, this one will be 30 when it is going that way.*0564

*The current of the river flows at 4 mph.*0570

*If we are looking at the speed of the boat and it is going against that river, probably this will be the boat - the current.*0575

*If we are looking at it going with the river, that will be the speed of the boat + the speed of the current.*0585

*They are helping each other out.*0590

*The only thing we do not know in here is the time, but I do know that the time was exactly the same for both of these situations.*0593

*Let us see what do we got here.*0603

*I will go ahead and create an equation for each of these.*0604

*x -4 × time = 10 and x + 4 × time = 30.*0607

*I know the times are exactly the same for each of these, let us solve them both for time.*0620

*This one I will go ahead and divide both sides by x - 4.*0628

*In this one I will divide by x + 4.*0631

*And I'm ready to develop that rational equation.*0638

*I will set each of these equal to each other since the times are equal to each other.*0641

*I have a rational equation then I can go ahead and try and solve.*0651

*10 ÷ x – 4 = 30 ÷ x + 4*0655

*To get through solving process, we find our common denominator.*0661

*I'm going to give x + 4 on the left side here and I will give x - 4 to the other side.*0668

*We will note that we made the denominators exactly the same.*0687

*We just need to focus on the tops of each of these.*0691

*Continuing on, you will distribute 10 and we will distribute 30.*0702

*10x + 40 = 30x -120*0710

*Subtracting a 10x from both sides will give us 40 = 20x -120 and let us go ahead and add 120 to both sides.*0720

*160 = 20*0739

*We can divide both sides by 20 and I have that x is equal to 8.*0745

*Let us make sure that it makes sense.*0754

*Some restricted values I have for my equations here I know that x cannot equal 4 and -4*0757

*and fortunately both that we have found for a possible solution is neither of those.*0766

*We can say the speed of the boat in still water would be 8 mph and then this guy is done.*0772

*Do not be afraid to use those tables from earlier to organize your information.*0790

*Joe and Steve operated a small roofing company and Mario can roof an average house alone in 9 hours.*0800

*Al can roof a house alone in 8 hours.*0809

*We want to know how long will it take them to do their job if they work together.*0812

*First we need to figure out the rates of each of them individually.*0818

*Let us go ahead and focus on Joe.*0821

*Joe can roof an average house alone in 9 hours.*0830

*Looking at just Joe we know that every hour he will get 1/9 of that house done.*0834

*Steve over here can roof the house in 8 hours.*0844

*He is working every single hour 1/8 of that house will be done.*0853

*We can put in that time on it.*0858

*Let x be the number of hours.*0861

*You have 1/9 × however many hours they work and 1/8 × qualified by however many hours Steve works.*0871

*We want to know how long it will take them to do the job if they work together.*0881

*We will take each of their work that they are doing and we will add them since they are working together,*0889

*we want to know when they will complete one job.*0895

*We have all of our information here and we can go ahead and try and solve this.*0899

*Our LCD would be 72.*0906

*We will multiply that through on all parts.*0911

*Doing some reducing 72 and 9 = 8, 72 and 8 = 9 and now we have an equation 8x + 9x = 72.*0923

*Combining together what we have on the left side this would be 17x is equal to 72*0942

*and we can divide both sides of that by 17 to get x = 72 ÷ 17.*0951

*It is looking pretty good.*0959

*That represents how long it will take them to work together.*0961

*If you want to represent that as a decimal, you could go ahead and take 72 ÷ 17.*0964

*When I did that I got about 4.24 hours, I did round it.*0971

*That gives me a better idea of how long it took them when working together.*0980

*Anytime when you are working with these types of problems, it should be less than any one of them working by themselves,*0985

*since they are helping each other out.*0992

*This one looks good.*0995

*In this last example we are going to look at a water tank that has two hoses connected to it.*0998

*Even though this is not a work problem you can see that we can set it up in much the same way.*1003

*Let us go ahead and give it a read.*1008

*The information that we have is that the first hose can fill the entire tank in 5 hours.*1011

*The second hose connected to this tank it can empty it in 3 hours.*1018

*If we start with a completely full tank and then we turned both of them open,*1023

*the question is how long will it take it to empty the entire tank?*1029

*Let us have an accrued picture of what we are dealing with here.*1034

*This would be our water tank and we have one hose that is going in and one hose that is going out.*1039

*The hose that is going in, it can fill the tank in 5 hours.*1054

*That means if we just leave it on every hour 1/5 of that tank will fill up.*1058

*I know its rate is 1/5.*1063

*If I look at emptying the tank it is a much smaller time to empty it, 1/3.*1068

*If I did have them both open I would know that the second one*1076

*would be able to empty the tank since it empties faster than the first one can fill it.*1080

*Let us see what that we can do to set this one out.*1086

*The first thing I want to consider was how much the tank is going to be empty every single hour.*1091

*Starting with the 1/3 I know that every hour that passes by the 1/3 of it will be emptied out.*1097

*x is the number of hours.*1106

*The 1/5 coming in is not emptying the tank but it is actually filling it back up.*1118

*We will say it is the opposite of emptying the tank by 1/5 and it will do that for every hour.*1125

*We want to know is when will one tank be completely empty.*1132

*We have a lot of similar components for this one and it often look like a lot of our work problems.*1137

*All we have to do is go ahead and solve it.*1145

*Our LCD here that would be 15.*1148

*Let us multiply all parts by 15 and see what that does.*1153

*15, 15 and 15.*1159

*Canceling out some extra stuff here I get 5x - 3x = 15.*1163

*Doing a little bit of combining on the left side I have 2x =15 or x = 15 ÷ 2 or 7 ½ hours.*1173

*Notice how in this case it is taking longer since both of them are open.*1190

*The reason for this is that they are not working with each other.*1194

*They are working against each other.*1198

*One was trying to fill the tank, and one is trying to empty the tank.*1199

*Use those bits of information you can find along the word problem to give you a little bit of intuition about your final solution.*1205

*In that way you can be assured that it does make sense in the context of the problem.*1212

*Thank you for watching www.educator.com.*1217

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